Question: $ D = \left[\begin{array}{rrr}-2 & -2 & -2 \\ 4 & 3 & 0\end{array}\right]$ $ v = \left[\begin{array}{r}3 \\ 1 \\ -2\end{array}\right]$ What is $ D v$ ?
Because $ D$ has dimensions $(2\times3)$ and $ v$ has dimensions $(3\times1)$ , the answer matrix will have dimensions $(2\times1)$ $ D v = \left[\begin{array}{rrr}{-2} & {-2} & {-2} \\ {4} & {3} & {0}\end{array}\right] \left[\begin{array}{r}{3} \\ {1} \\ {-2}\end{array}\right] = \left[\begin{array}{r}? \\ ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ v$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ v$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ v$ , and so on. Add the products together. $ \left[\begin{array}{r}{-2}\cdot{3}+{-2}\cdot{1}+{-2}\cdot{-2} \\ ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ v$ and add the products together. $ \left[\begin{array}{r}{-2}\cdot{3}+{-2}\cdot{1}+{-2}\cdot{-2} \\ {4}\cdot{3}+{3}\cdot{1}+{0}\cdot{-2}\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{r}{-2}\cdot{3}+{-2}\cdot{1}+{-2}\cdot{-2} \\ {4}\cdot{3}+{3}\cdot{1}+{0}\cdot{-2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{r}-4 \\ 15\end{array}\right] $